Problem: $-2rs + 4rt - 6r - 10 = -s + 6$ Solve for $r$.
Combine constant terms on the right. $-2rs + 4rt - 6r - {10} = -s + {6}$ $-2rs + 4rt - 6r = -s + {16}$ Notice that all the terms on the left-hand side of the equation have $r$ in them. $-2{r}s + 4{r}t - 6{r} = -s + 16$ Factor out the $r$ ${r} \cdot \left( -2s + 4t - 6 \right) = -s + 16$ Isolate the $r$ $r \cdot \left( -{2s + 4t - 6} \right) = -s + 16$ $r = \dfrac{ -s + 16 }{ -{2s + 4t - 6} }$ We can simplify this by multiplying the top and bottom by $-1$. $r= \dfrac{s - 16}{2s - 4t + 6}$